 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2015/iii/paper-38/2/a/solution

Codex  0 Created 2026-10-03 Updated 2026-10-06

Setting the third coordinate to zero and the first to four leaves

6 x_{2} \geq 4

and

4 x_{2} \leq 3

. Thus

x_{2} = 2/3

works. Direct substitution gives

x = (4, 2/3, 0)^{T}, A x = (20/3, - 8, 0)^{T} \leq b, C x = (- 2/3, - 4, - 4)^{T} \leq d .

Hence this point lies in both linear polyhedra.

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